The following is supposed to be "clear" according to Kueker, but I could not see why. Can anyone help?

Let $A$ be a countable structure with uncountable many automorphisms. Then for every $\vec{a}\in A^{<\omega}$, $(A,\vec{a})$ has a non-trivial automorphism, i.e. there exists some $f:A\rightarrow A$ such that $f\neq id$ and $f(a)=a$, for all $a\in \vec{a}$.

1 Answer
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Let $\vec{a} \in A^{<\omega}$. There are uncountably many automorphisms $f$ and only countably many possible values for $f(\vec{a})$, so there must be two different automorphisms $f_1$ and $f_2$ with $f_1(\vec{a}) = f_2(\vec{a})$. Then $f_2^{-1} \circ f_1$ is a nontrivial automorphism fixing $\vec{a}$.

In fact, there are uncountably many automorphisms moving $\vec{a}$ the same way, so fixing one and composing the others with its inverse gives uncountably many nontrivial automorphisms fixing $\vec{a}$.